Video: Finding the Derivative of a Polar Equation

Given a polar curve defined by π‘Ÿ = 𝑓(ΞΈ), form an expression for the slope of the curve d𝑦/dπ‘₯ in terms of πœƒ and 𝑓.

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Video Transcript

Given the polar curve defined by π‘Ÿ is equal to 𝑓 of πœƒ, form an expression for the slope of the curve the derivative of 𝑦 with respect to π‘₯ in terms of πœƒ and 𝑓.

The question gives us a polar curve defined by the equation π‘Ÿ is equal to 𝑓 of πœƒ. And it’s asking us to form an expression for the slope of the curve. That’s the derivative of 𝑦 with respect to π‘₯. And it wants this in terms of πœƒ and 𝑓. To do this, we recall the chain rule, which tells us that the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to πœƒ multiplied by the derivative of πœƒ with respect to π‘₯. So, this is the first step. This gives us an equation for our slope function d𝑦 by dπ‘₯.

We now recall our polar equations, 𝑦 is equal to π‘Ÿ sin πœƒ and π‘₯ is equal to π‘Ÿ cos πœƒ. We recall the question is asking us to write our slope function in terms of only πœƒ and 𝑓. So, it would be a good idea to rewrite any equations in terms of πœƒ and 𝑓 where possible. Luckily, we can use the substitution π‘Ÿ is equal to 𝑓 of πœƒ to rewrite these polar equations.

At this point, we could calculate the derivative of 𝑦 with respect to πœƒ without any problems. However, we can’t directly calculate the derivative of πœƒ with respect to π‘₯, since we’re given π‘₯ as a function of πœƒ. We can get around this by recalling that multiplying by the derivative of πœƒ with respect to π‘₯ is the same as dividing by the derivative of π‘₯ with respect to πœƒ. So, we can find an expression for our slope function d𝑦 by dπ‘₯ by first calculating d𝑦 by dπœƒ and then dividing this by dπ‘₯ by dπœƒ. So, to calculate the derivative of 𝑦 with respect to πœƒ, we differentiate 𝑓 of πœƒ multiplied by sin πœƒ with respect to πœƒ.

We can differentiate this by recalling the product rule, which tells us the derivative of the product of two functions 𝑒 and 𝑣 is equal to 𝑒 prime multiplied by 𝑣 plus 𝑣 prime multiplied by 𝑒. So, we first differentiate our function 𝑓 of πœƒ to get 𝑓 prime of πœƒ and multiply this by sin πœƒ. Then, we differentiate sin πœƒ, which we recall is cos πœƒ. And we multiply this by 𝑓 of πœƒ. We can then do the same to calculate the derivative of π‘₯ with respect to πœƒ. So, that’s the derivative of 𝑓 of πœƒ multiplied by the cos of πœƒ with respect to πœƒ.

We differentiate 𝑓 of πœƒ to get 𝑓 prime of πœƒ and then multiply this by the cos of πœƒ. And we add onto this the derivative of the cos of πœƒ, which we recall is negative sin πœƒ. And we multiply this by 𝑓 of πœƒ. We’re now ready to substitute this into our equation for our slope function. We have that d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ, divided by dπ‘₯ by dπœƒ. Substituting the equation we got for the derivative of 𝑦 with respect to πœƒ gives us the numerator of 𝑓 prime of πœƒ multiplied by sin πœƒ plus 𝑓 of πœƒ multiplied by the cos of πœƒ.

And then, substituting the equation we got for the derivative of π‘₯ with respect to πœƒ gives us a denominator of 𝑓 prime of πœƒ multiplied by the cos of πœƒ minus 𝑓 of πœƒ multiplied by the sin of πœƒ. And so, we have found the formula for our slope function d𝑦 dπ‘₯ in terms of the function 𝑓 and πœƒ. And it’s fine to just quote this. We don’t need to derive this every time. And so, in conclusion, we have shown that if we are given a polar curve defined by the equation 𝑓 of πœƒ. Then, we can rewrite the slope function the derivative of 𝑦 with respect to π‘₯ as 𝑓 prime of πœƒ multiplied by sin πœƒ plus 𝑓 of πœƒ multiplied by the cos of πœƒ all divided by 𝑓 prime of πœƒ multiplied by the cos of πœƒ minus 𝑓 of πœƒ multiplied by the sin of πœƒ.

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